Given a positive integer $n$, consider $f_n = -\min_{|z|=1} \Re \sum_{i>n} \frac{z^i e^{-i/n}}{i}$. What can be said about the growth of $f_n$? How large can it get?

Taking maximum instead of minimum, the answer is obvious as the maximum is attained at $z=1$ and this becomes a real analysis problem (which in particular implies $f_n$ is bounded, but perhaps $f_n$ tends to $0$ at some rate).

An approach suggested by a friend is to express $\sum_{i>n} t^i/i$ in terms of a Cauchy integral related to $-\ln (1-\zeta t)$ and then perhaps apply a saddle point estimate (uniformly in $|t|=e^{-1/n}$) -- but I wasn't able to carry this out.


The key is the following integral representation:
\begin{equation} \begin{aligned} s_n(z)&:= \sum_{j>n} \frac{z^j e^{-j/n}}j \\ &=\sum_{j>n} z^j e^{-j/n} \int_0^\infty du\,e^{-ju} \\ &=\int_0^\infty du\,\sum_{j>n} z^j e^{-j/n}e^{-ju} \\ &=\frac{z^{n+1}}{e^{1+1/n}}\,\int_0^\infty du\,\frac{e^{-(n+1)u}}{1-z e^{-u-1/n}} \\ &=\frac{z^{n+1}}{e^{1+1/n}}\,\int_0^\infty dv\,\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}. \end{aligned} \tag{1} \end{equation} Next, with $n\ge1$, $|z|=1$, and $v>0$, \begin{equation*} |1-z e^{-(v+1)/n}|\ge1-e^{-(v+1)/n}\ge1-\frac1{1+(v+1)/n}=\frac{v+1}{n+v+1}, \end{equation*} whence \begin{equation*} \Big|\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}\Big| \le2e^{-v}. \tag{2} \end{equation*} Also, for all real $t$ and all real $v>0$, \begin{equation*} \Big|\frac{e^{-v}}{1+v-it}\Big|\le e^{-v}. \tag{3} \end{equation*}

Suppose now that $n\to\infty$ and $z=e^{it/n}$ for some real number $t$ possibly varying with $n$ but staying bounded. Then \begin{equation*} \frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}=\frac{e^{-v}}{1+v-it}+o(1) \end{equation*} for each real $v>0$. So, by (1), (2), (3) and dominated convergence, \begin{align*} s_n(z)&= \frac{e^{it}}e\,\int_0^\infty dv\,\frac{e^{-v}}{1+v-it}+o(1)=\Gamma(0,1-i t)+o(1). \end{align*}

On the other hand, if for $z$ with $|z|=1$ it is not true that $z=e^{it/n}$ for some real number $t$ possibly varying with $n$ but staying bounded, then without loss of generality $n(1-z)\to\infty$, whence \begin{equation*} n|1-z e^{-(v+1)/n}|\ge n|1-z|-(v+1)\to\infty \end{equation*} for each real $v>0$. So, by (1), (2), and dominated convergence, $|s_n(z)|\to0$.

So, \begin{equation*} -f_n=\min_{|z|=1} \Re s_n(z)\to\mu, \end{equation*} where \begin{equation*} \mu:=\min_{t>0}r(t),\quad r(t):=\Re\Gamma(0,1-i t). \end{equation*} (According to Mathematica's numerics, $\mu=r(t_*)=-0.12686\dots$, where $t_*=2.0287\dots$.)

  • $\begingroup$ Thank you Iosif, this is clear and helpful. $\endgroup$ – Ofir Gorodetsky Apr 8 at 14:18

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